Infinitesimal transformations and Poisson brackets

[JonnyMaddox]

Hello, I want to understand how bracket operations in general are related to symmetry and infinitesimal transformations (in hindsight of quantumfieldtheory), so I calculated an example with a particle that is moving on a circle with a generic potential.
(I used simple polar coordinates in two dimensions)

H(r,p_{r})= \frac{p^{2}_{r}}{2m}+V(r)
H(\phi, p_{\phi})=\frac{p^{2}_{\phi}}{2mr^{2}}+V(\phi)

Now I know that if you take the Poisson bracket with the Hamiltonian you just get the infinitesimal transformation in time right? So
\{r,H(r,p_{r})\}= \frac{p_{r}}{m}
\{\phi,H(\phi,p_{\phi})\}= \frac{p_{\phi}}{mr^{2}}

But what if I want to do an infinitesimal transformation of the r and \phi coordinates? I know that the generator of translations is just the momentum, and that the generator of rotations is angular momentum. How would I do that with the Poisson bracket in this case? And for example when I do an infinitesimal transformation with the radius r, what does that mean? Is it that the radius is infinitesimally transformed, or is it more like a global translation where the whole system is somehow translated? Similarly with the angle \phi, is it that the angle is locally changed, or is it that the "whole" system is rotated?

 

[ch3cooh]

Now I know that if you take the Poisson bracket with the Hamiltonian you just get the infinitesimal transformation in time right? So

Are you referring to this formula
$$\frac{df}{dt} = \{f, H\} + \frac{\partial f}{\partial t} $$
As for infinitesimal transformation, it is just the radius $r$ changes. For example, we may have a infinitesimal canonical transformation
$$Q = q + \alpha G(q,P)$$
where $\alpha$ is infinitesimally small and $G$ is the generator.